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Simora [160]
1 year ago
8

Really need help guys! Love you if you can solve them wiz steps!!!!

Mathematics
1 answer:
Marta_Voda [28]1 year ago
5 0

Answer:

Step-by-step explanation:

x^2+y^2-8x+4y+4=0\\\\a)\\\\x^2-8x+y^2+4y+4=0\\\\x^2-2*x*4+(y^2+2*y*2+2^2)=0\\\\x^2-2*x*4+4^2+(y+2)^2=4^2\\\\(x-4)^2+(y+2)^2=4^2\\

Hence,

The radius of the circle is 4 units, coordinates of its centre are (4,-2).

b)\\\\y=0\\x^2+0^2-8x+4*0+4=0\\x^2-8x+4=0\\a=1\ \ \ \ b=-8\ \ \ \ c=4\\D=(-8)^2-4*1*4\\D=64-14\\D=48\\\sqrt{D}=\sqrt{48} \\ \sqrt{D}=\sqrt{16*3} \\\sqrt{D}=\sqrt{4^2*3}  \\\sqrt{D}=4\sqrt{3}  \\\displaystyle\\x=\frac{-(-8)б4\sqrt{3} }{2*1} \\\\x=\frac{8б4\sqrt{3} }{2} \\x_1=4-2\sqrt{3} \\x_2=4+2\sqrt{3}

c)\\\\A(6,2\sqrt{3}-2)\\ (x-4)^2+(y+2)^2=4^2\\(6-4)^2+(2\sqrt{3} -2+2)^2=16\\2^2+(2\sqrt{3} )^2=16\\4+2^2*(\sqrt{3})^2=16\\ 4+4*3=16\\4+12=16\\16\equiv16

d)\\A(6,2\sqrt{3}-2)}\\\sqrt{3} x+3y=\\\sqrt{3}(6)+3(2\sqrt{3} -2)=\\ 6\sqrt{3}+6\sqrt{3} -6=\\ 12\sqrt{3}-6

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FrozenT [24]

Answer: the height of the pole is 16.1 m.

Step-by-step explanation:

The scenario is illustrated in the attached photo.

x represents the height of the pole.

y represents the distance from the foot of one stabilizing wire to the foot of the pole.

30 - y represents the distance from the foot of the other stabilizing wire to the foot of the pole.

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Tan θ, = opposite side/adjacent side.

Considering triangle ACD,

Tan 60 = x/y

x = ytan60 = y × 1.732

x = 1.732y- - - - - - - - -1

Considering triangle BCD,

Tan 38 = x/(30 - y)

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